33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

Telegram @unacademyplusdiscounts
124 NEET-AIPMT Chapterwise Topicwise Solutions Physics
31. (b) : Resultant amplitude = 2a (1 + cosf) = a
1 2π
∴ ( 1+ cos φ) = 1/ 2;cos φ=− ; φ = .
2 3
32. (b) : Given : x = acos(wt + d)
and y = acos(wt + a)...(i)
where, d = a + p/2
\ x = acos(wt + a + p/2) = –asin(wt + a)...(ii)
Given the two waves are acting in perpendicular direction
with the same frequency and phase difference p/2.
From equations (i) and (ii),
x 2 + y 2 = a 2
which represents the equation of a circle.
33. (c) : Frequency (u) = 800 Hz
As the pipe is closed at one end, so
l 3 – l 2 = l 2 –l 1 = λ = 21. 5 cm
2
\ l = 43.0 cm
v
As υ = ⇒ v = υλ λ
800 × 43
−1
∴ v = = 344 ms
100
34. (b) : The velocity of sound in air at 27°C is
v = 2(u) [L 2 – L 1 ]; where u = frequency of tuning fork
and L 1 , L 2 are the successive column length.
\ v = 2 × 320[73 – 20] × 10 –2
= 339.2 m s –1 ≈ 339 m s –1
35. (a) : For closed organ pipe, third harmonic is 3 v
4l .
v
For open organ pipe, fundamental frequency is
2 l′
Given, third harmonic for closed organ pipe
= fundamental frequency for open organ pipe.
3v
v
4l 2l ∴ = ⇒ l′ = = ;
4l
2l
′ 3 × 2 3
where l and l′ are the lengths of closed and open organ
pipes respectively.
2
∴ l′= × 20
= 13. 33 cm
3
36. (a) : Nearest harmonics of an organ pipe closed at
one end is differ by twice of its fundamental frequency.
\ 260 – 220 = 2u, u = 20 Hz
37. (b) : Second overtone of an open organ pipe
v
( Thirdharmonic ) = 3× υ 0
′ = 3×
2 L′
First overtone of a closed organ pipe
v
( Thirdharmonic ) = 3× υ 0 = 3×
4 L
According to question,
3 υ 0′ = 3 υ v
0 ⇒ 3 × 3 2 ′ = × v
L 4 L
\ L′ = 2L
38. (a) : From figure,
First harmonic is obtained at
λ
l = = 50 cm
4
Third harmonic is obtained for resonance,
3λ
l′ = = 3× 50 = 150 cm.
4
39. (b) : For a string fixed at both ends, the resonant
frequencies are
nv
υ n = where n = 123 , , ,.....
2L
The difference between two consecutive resonant
frequencies is
( n+
1)
v nv v
∆υn = υn+1
− υn
= − =
2L
2L
2L
which is also the lowest resonant frequency
(n = 1).
Thus the lowest resonant frequency for the given string
= 420 Hz –315 Hz = 105 Hz
40. (a) : For closed organ pipe, fundamental frequency
v
is given by υ c = 4l
For open organ pipe, fundamental frequency is given by
v
υ o = 2 l ′
2 nd overtone of open organ pipe
3
υ′ = 3υ0 ; v
υ′ =
2l
′
According to question, u c = u′
v 3v
= ; l′ = 6l
4l
2l
′
Here, l = 20 cm, l′ = ?
\ l′ = 6 × 20 = 120 cm
41. (d) : Fundamental frequency of the closed organ
pipe is
υ= v 4 L
Here, v = 340 m s –1 , L = 85 cm = 0.85 m
−
∴ υ = 340 1
ms
= 100 Hz
4 × 085 . m
The natural frequencies of the closed organ pipe will be
u n = (2n – 1)u = u, 3u, 5u, 7u, 9u, 11u, 13u,...
= 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz,
1300 Hz,... and so on
Thus, the number of natural frequencies lies below
1250 Hz is 6.